5
$\begingroup$

Define an "integer 3-relationship" as a function $f(a,b,c)$ of three integer variables, together with the condition that this function must equal zero. Two examples would be the "Pythagorean triplet relationship" with $f(a,b,c) \equiv c^2 - (a^2 + b^2) $ and the "near-miss Pythagorean triplet relationship" with $f(a,b,c) \equiv c^2 +1 - (a^2 + b^2) $.

For any given integer 3-relationship based of some $f$, there may be many triplets that satisfy the condition. For example, any triplet $$ \{k(m^2 - n^2), \,2k m n, \, k(m^2+n^2)\}\,\, (k,m,n \in \Bbb N) $$ satisfies the Pythagorean triplet relationship.

In some cases, there may be a chain of $n$ triplets $$ \{a, b, c\},\, \{b, c, d\},\, \{c, d, e\},\, \cdots $$ such that each of these triplets satisfies the same integer 3-relationship based of some $f$, and the last two entries in each triplet match the first two entries in the next. For example, with $f(a,b,c) = c^2 +1 - (a^2 + b^2)$, the triplets $\{7,\, 11,\, 13\}, \{11,\, 13,\, 17\}$ form a 2-chain because $$ 7^2 + 11^2 = 13^2+1 \\ 11^2 + 13^2 = 17^2 + 1 $$ I am concerned with looking for chains of Pythagorean triplets.

Prove that (other that trivial chains with $a$ or $b$ zero) there are no 2-chains of Pythagorean triplets (or disprove that conjecture by providing an example Pythagorean 2-chain).


I think I have a proof along the lines of starting with a "primitive" 2-chain (with no common factor in $a$ and $b$), applying the general form of Pythagorean triplets given above, and re-distributing the factors in $2 m n$ to form the $2 r s$ of the next triplet, so that $ 2mn = 2rs$ and $m^2 + n^2 = r^2 - s^2$. At that point I can solve a quadratic equation for one of the factors and the condition that the discriminant be a perfect square leads to the construction of a different 2-chain with smaller numbers -- and reduction ad absurdium applies.

But my proof is not what I consider solid...

$\endgroup$
0
$\begingroup$

For a chain of Pythagorean triplets we have$$b^2=c^2-a^2$$and$$b^2=d^2-c^2.$$ Hence $a^2,c^2$ and $d^2$ are in arithmetic progression with common difference $b^2.$

A proof that three squares in arithmetic progression cannot have a square as the common difference was given by Fermat and is known as Fermat's right triangle theorem. (The proof is included on the linked page.)

A number that can be a difference of three squares in AP is known as a congruum and it is known that these numbers are those that are exactly four times the area of a Pythagorean triangle. Hence, if a congruum is a square, then the area of the Pythagorean triangle must also be square.

Fermat's proof relies on showing that if there exists a Pythagorean triangle with an area that is a square number then a smaller such Pythagorean triangle must exist.

$\endgroup$
  • $\begingroup$ Apparently this is the one and only complete number theory proof found in Fermat's surviving work and was written in the margin of the same book that the Last Theorem comment was found. $\endgroup$ – nickgard Jun 21 '18 at 9:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.